Problem: What is the slope of the line tangent to $f(x) = x^{2}+x-6$ at $x = -1$ ?
Explanation: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{((x+h)^{2}+x+h-6) - (x^{2}+x-6)}{h}$ $ = \lim_{h \to 0} \frac{(x^{2}+2x h+h^{2}+x+h-6) - (x^{2}+x-6)}{h}$ $ = \lim_{h \to 0} \frac{x^{2}+2(x h)+h^{2}+x+h-6-x^{2}-x+6}{h}$ $ = \lim_{h \to 0} \frac{2(x h)+h^{2}+h}{h}$ $ = \lim_{h \to 0} 2x+h+1$ $ = 2x+1$ $ = (2)(-1)+1$ $ = -1$